Johann Marinsek
2014
[email protected]
Metallic bonding:
Electrons don’t provide the glue holding solids together
Genuine metallic lattices consist of elastic rods made of hydrogen atoms
Metallic bonding is not due to forces between alleged free valence electrons and
atomic core ions. There is no natural law that governs the process how and how
many valence electrons are detached from atoms in order to do their job as bonding
electrons. Electrons are picked on demand.
Electrons are also neither the agents for electrical
and thermal conduction nor for specific heat. The
existence of free electrons in metals is not proven by
experiment. Alternative model to electron bonding:
Metallic lattices possess
• elastic rods that consist of hydrogen atoms and
• nodes that contain hydrogen atoms.
Structural elements explain stabilities of different
crystal structures.
In order to explain diverging physical properties of
bcc metals one must differentiate between two bcc
crystal structures:
fcc: space-filling-parallel-epiped
• α-bbc (“soft” bcc) for alkali metals and
• β-bbc (”hard” bcc) for alkali earth and transition
metals.
Stable building blocks of fcc lattices are parallel-epipeds containing an octahedron
and two regular tetrahedrons.
There are 3 modes of bonding rods:
single (|) H, double parallel (||) H’s, two H’s in series (——) and double H’s in series
(= =).
In the proposed genuine lattice model double bonds (2 rods) are possible to explain
different physical properties for two metals with almost the same lattice constant a0
and mass number A whereas the ruling doctrine cannot assume two different forces
for charges and atom distances of nearly the same magnitude.
Physical properties of metals depend mainly on
• crystal system,
• bonding mode and
• primitive unit cell dimensions.
Effective bond lengths L of unit cells depend on mass number A. The bigger mass
the shorter effective rod length. Melting occurs when bonds reach failure due to
resonance. Resonance frequency depends on 1/L2. Also specific heat depends on
1/L2.
1
Unit cells of metals are not isotropic. Due to anisotropy some physical properties
(elasticity moduli E, G, B; electrical conductivity; sound velocities) possess different
values depending on direction. For cubic crystals the ratio of longitudinal to
transversal sound velocities is 2.0.
Controversy on metallic bonding
Einstein: … free electrons do not exist in metals at all.
versus
M. L. Cohen: The Electron’s Central Role
Electrons provide the glue holding solids together, and hence they are central in
determining structural, mechanical and vibrational properties. [coh]
This is the basic ontology of Solid State Quantum Mechanics!
Anomalies refute bonding theory of quantum mechanics
Quantum mechanics beyond remedy…
The model of valence electrons that provide the glue holding solids together lacks any
causality.
The lack of causality concerns the baseless choice of the number of valence electrons. An
example may elucidate this choice. The claim is that valence electrons are ready to serve
for bonding because they are loosely bound to the atomic nuclei. So Cr and V with
valency 6 and 5 possess 5 and 6 electrons as bonding agents, respectively.
Potassium (K) for example is modest; for bonding the unique valence electron
suffices…Comparing ionization energies we see that for the first 6 electrons of Cr the
following energies [eV] to detach the electrons are necessary:
Cr:
6.8;
16.5;
31.;
49.2;
69.5;
90.6
[eV]
82.6;
99,4
[eV]
Now take K, the corresponding energies are:
K:
4.3;
31.6;
45.8;
61.;
If 90.6 [eV] suffice to detach the 6th electron of Cr, then the 5th electron of K is also
detachable ( 82.6 [eV] necessary). So the question remains: Why only one electron of K
must do the job whereas for Cr 5 electrons do? Quantum theorists don’t mention the
cause; electrons must have a brain to decide what is to do… for the weak bond of
potassium only one electron can build up the forces that hold the lattice together…
Concept of valence electrons worthless for metallic bonding
The concept of valence has its origin in chemistry. Concerning metallic bonding
Wikipedia declares on valence electrons:
Because the number of valence electrons which actually participate in chemical reactions
is difficult to predict, the concept of the valence electron is less useful for a transition
metal than for a main group element...
Concerning metallic bonding the only interesting problem is the range of ionization
energies for possible detachments of electrons that serves as the glue to hold metals
together. Nowhere quantum mechanics makes mention of such a range. Moreover,
ionization energies for the 6th, the 7th and so one of Ir are missing.
2
How many valence electrons possess Pt and Ir? 5 or 6? As you like! For
periodictable.com both possess 6 valence electrons. But then the electrostatic forces
between electrons and atomic ions would be equal for both, Ir and Pt. But elasticity
modulus E of Ir is about three (3!) times greater than for Pt.
Now take 5 valence electrons for Pt and 6 for Ir. The ratio of electrostatic forces for Ir/Pt
would be 62/52 = 1.44. (For roughly the same lattice constant a0.)
A theory in trouble... no wonder that Pauling picks number of valence electrons on
demand. See Appendix.
Recall that the model of forces between valence electrons and atomic ions calculates a
promiscuity of forces that does not exist. Therefore the theoretical elasticity values Cij are
not the elasticities of a genuine lattice that consists of rods and nodes.
Concerning crystal structure, atomic weight, number of protons, lattice constant, both,
Ir and Pt are only marginally different or even equal. Both, Ir and Pt are paramagnetic.
Note that both elements, Ir and Pt, possess (according to periodictable.com) the same
number of valence electrons, namely 6. So the assumed virtual force lattices due to
electrostatic forces between electrons and atomic core ions don’t differ. Ir as well as Pt
atomic cores would be 6+ ions.
Therefore physical and electrical properties of Ir and Pt should be approximately equal –
they are not:
Fermi energies are approximately the same for both Ir and Pt, but electrical conductivity
of Ir is two times greater than for Pt!
When the alleged free electrons of Ir and Pt should perform the same electricity induced
movements (because Ir and Pt lattices don’t differ significantly), then why does electrical
conduction differ by 200%?
According to measurements Young’s modulus E of Ir is about 3-fold greater than for Pt.
A double bond (||) for Ir instead of a single one (|) for Pt can explain higher E- modulus
and Vickers hardness of Ir.
The current model fails in this respect because the forces between Ir- and Pt- ions are
roughly equal! Therefore E- moduli should be equal, they are not…
When Ir lattice rods are twin rods || then the buckling load is a multiple of the buckling
load of Pt that has single rods |.
Smaller thermal expansion of Ir is due to double bonds.
Stronger bonds and higher melting point of Ir could be explained by QM only with more
delocalised valence electrons of the Ir lattice.
QM is restricted to one and only one force between electrons and core ions. By no
means can one duplicate or triplicate this force for Ir in order to fit the data!
To augment arbitrarily the number of valence electrons is not possible…
Take a look at the table:
f
c
c
B
o
n
d
Mass
#
A
Ir
||
192
Pt
|
Ir/
Pt
2
Proton
#
Valency
z
Density
g/cm3
n=
z×ρ/A
electrons
Lattice
77
6 (5)
195
78
0.98
0.99
102/a02
GPa
E
Pois
son
γ
Melting
K
22.5
0.7
3.84
6.78
525
0.26
6
21.5
0.66
3.92
6.51
169
1
1.05
1.06
0.98
1.04
3.1
a0
Sound Therm
cL
cond.
103 m/s W/
(mK)
Therm
exp.
1/106K
2683
Electr.
conduct.
106
S/m
19.7
5.35
150
6.4
0.38
2045
9.43
3.83
72
8.8
0.68
1.3
2.1
1.4
2.08
0.73
[Å]
Legend: cL = longitudinal sound velocity
3
Comparison of physical properties of Ir and Pt:
Maximum deviations show moduli E.
Young’s modulus E = (Force/Area) (Length/ΔL).
Analogue for Young’s modulus:
If a spring possesses a spring constant E = 1 and
we add an identical spring parallel, than this
composition shows E = 2. For equal large unit cells
double bonding means double elasticity modulus E.
E depends on the number of bonds and on 1/a02, where a0 is the area of the face of the cubic
unit cell. Regarding the Ir/Pt ratio for E-moduli we must take into account the smaller
(about 4%) unit cell area of Pt. Comparing E-moduli of Ir and Pt the ratio is not 525/169 =
3.1 (according to measured values) but 525/169+4% = 2.98.
But the expected theoretical value is about 2.0; below we cite examples where this factor
occurs!
Below we will explain why measured values are not exactly representative for E-moduli:
Real macroscopic pieces of metals are polycristalline aggregates. This means a
conglomerate of metal grains that possess different directons of physical properties like
elasticity, electrical conduction. Polycrstalline metal pieces are roughly isotropic.
Single crystals don’t exist. To do experiments there are agglomerates of single crystals
necessary, at least several milimeters large. Such experiments show anisotropy due
to crystal structure. Single crystals of metals possess not one unique E- modulus but 3 of
them!
Pay attention that we have not pure metals and that we have isotope mixtures. For some
metals measured E- moduli show a wide range… Example Os: E-modulus. 548-570 GPa
“Isotope alloys” ; impurities
Impurities influence E- moduli: 99.99% Pt: E = 169; 99.9 % Pt, E = 177 GPa!
Note that Pt is a mix of 192, 194, 195, 196 and 198 isotopes (A = mass number).
Source: J. Merker et.al.:Platinum Metals Rev. 2001, 45, (2)
An alloy of 99% Cu and 1% Ag shows a significant increase of electrical conductivity
compared with pure Cu! So it is probable that pure isotopes of a metal and its isotope mix
possess not the same physical properties. It is necessary to separate the isotopes and to
measure the properties of the isotopes. A striking example for the isotope effect is carbon.
Naturally occurring diamonds consist not of 100% 12C but contain 1% 13C. For a pure 12C
diamond thermal conductivity for example shows an increase of > 50%!
(D. Suter, Festkörperphysik, e3.physik.uni-dortmund.de/~suter/Vorlesung/.../4_Phononen.pdf)
Isotope shift: 12C and 13C show different Raman frequencies, for example.
Melting
a
supporting
node
The collapse mode of melting is the resonance
L
mode
catastrophes for lattice rods when temperature or
shock wave caused oscillations reach Eigen
L = effective rod length
frequencies.
The Eigen frequencies of bars depend predominantly on 1/L2.
Effective bond length L is ca. the same for Ir and Pt.
4
Resonance frequency depend also on supporting modes. Supporting modes for single | or
double bonding || are probably different.
When the thermal expansion of Pt is greater than for Ir, the melting point of Pt decreases,
for Ir it increases…
So the ratio of melting temperatures for Ir/Pt is not 1 (as one would expect) but 1.3. But
below we see for a similar case that the melting point ratio of Cr/V is ∼ 1!
Regarding the build-up process of metal lattices, QM never explained how and where the
liberation of bound valence electrons occurs…
Furthermore we shall explain that free valence electrons simply do not exist!
Again: Physical properties of metal lattices are due to crystal structure and bonding
mode only.
Predominant is the crystal system (bcc, fcc etc).
Influencing factors are
1: the character of bonding: single (|) H, double (||) H’s;
two H’s in series(——), double H’s in series (= =);
2: the lattice constant a0;
3: effective length of lattice bars L.
The dimensions of lattice nodes determine L. For current theory of metallic bonding the
dimensions of the atoms have no influence on forces because forces depend only on the
distance between atomic centres.
Cr compared to V
The table below contains a comparison of the bcc metals Cr and V. As before for Ir and
Pt, atomic weight, number of protons, lattice constant, for both, Cr and V are only
marginally different or even equal. (There are 4 naturally occurring stable isotopes for Cr, 84% are
52
Cr. Vanadium consists of two isotopes, 99.75 % are 51V.)
According to periodictable.com Cr possesses 6 valence electrons, V its 5.
(Other sources propose 2 -5 valence electrons for V and 2-6 electrons for Cr.)
So forces in the Cr-lattice are greater, but not more than twofold. Note that the
elasticity moduli ratio of Cr/V = 2.2! But specific heats per volume and melting points
are roughly the same… QM cannot explain this behaviour…
bcc
B
o
n
d
a0
Cr-52
Sound
103m/s cL/cT
cL
Therm
Cond
W/
(mK)
Thermal
exp
8.74
94
4.9
2180
5
31
8.4
1.
1.75
3.03
0.6
spec
heat
J/
cm3K
Mel
ting
K
El.
cond
106
S/m
1.65
3.19
2180
6.09
2.2
3
1.3
0.75
1.1
[Å]
Valence
E
GPa
Brinell
MPa
Pois
-son
γ
||
2.91
6
279
1120
0.21
6.63
V-51
|
3.03
5
128
628
0.37
Cr/V
2
0.96
1.2
2.2
1.8
0.57
Legend: cL = longitudinal sound velocity; cT = transversal sound velocity
Remark to all tables: Data are from [per], [che], [rau],[roy]. Note that vague data arise from impurities and
isotope mixtures of metals! There are different values for physical properties listed.
If one assumes that the bonding of Cr consists of 2 elastic rods || , whereas the lattice V
possesses only one rod |, than the larger E modulus of Cr appears plausible.
Now one must ask the question why the metals Cr and V possess different stiffness but
not different melting points. For QM with its electrons that should glue the atom ions
together this is an insoluble problem one.
5
When we think at elastic rods as bonds that undergo a break down due to a resonance
frequency, then a second rod cannot hinder the failure of the lattice due to resonance
because it possesses the same resonance frequency!
So melting points of Cr and V are equal, independent of the different bonding: || for Cr
versus | or V. Different supporting modes of bonds and different thermal expansion value
obviously equalise each other.
Similarly, specific heat absorption does not depend on the number of bonds…
Obviously, specific heats of V and Cr don’t differ significantly.
Smaller thermal expansion of Cr is due to double bonds.
Metallic bonding according to QM
Cohen’s QM ontology, continued:
Thomson’s plum pudding model resembled our modern picture of jellium with a positive
smeared out background representing the ions and these electrons existing in this
background. Unlike jellium where the electrons are smeared out, Thomson’s electrons are
plums. Hence, the essential difference is that the electrons in the jellium model are treated
quantum mechanically and despite the fact that they can be excited out of the metal and
look like Thomson’s plums, inside the metal they are itinerant.
According to the prevailing quantum theory, metallic bonding is the interplay of free
valence electrons (that constitute the so-called electron cloud) with the ionized cores of the
lattice atoms. The electrons allegedly drift in a random manner through the lattice of ion
cores. The bonding model explains metals as a lattice of core ions held together by a gas of
free electrons due to electrostatic distant forces that act through a vacuum!
Of course, there is no genuine real lattice because lattice rods are missing!
The tramping and oscillating electron gas cannot hold together the oscillating cores ions
(that are repelling each other) in order to form a stable lattice! Metals are stable to such
an extent that they are malleable! A calculated proof of elastic equilibrium does not
involve malleability…
Survey of QM bonding theory
anomalies
Erroneous presuppositions of QM
metallic bonding
• The model is a quasi-continuous theory of
elasticity because it supposes a dense web of
forces. Indeed the claim is that metals are
microscopically anisotropic.
• The existence of electrons in electron shells
around a nucleus that is made out of neutrons
and protons is not proven. To save phenomena
does not prove a model.
6
• The number of free bonding electrons is chosen arbitrarily in order to get adequate force
magnitudes for equilibrium. No limits for the number of needed electrons are deduced
from a theory.
Concerning the increasing liberation energy for more electrons there is no scrupulosity.
According to QM solidifying means liberation of electrons. This process is vaguely
described as follows: When atoms are closely packed, the wave functions of the outermost
(bonding) electrons overlap. Then these electrons can abandon the local potentials and are
„free“.
Of course the electron cloud is not observable. To save phenomena was not the question
but to save a theory!
• QM is vacuum physics. For electromagnetic waves there is no carrier. The space (that is
a relation, not substance) itself is not waving, this is a categorical mistake. There is no void
but a plenum, namely a dielectric that consists of electrons and positrons. Faraday argued
therefore that distant forces are unreal. QM assumes levitating electrons and ions in the
void (in vacuo).
• But even if we assume tentatively the existence of distant central forces, a stable
equilibrium of interacting repulsive and attractive electrostatic forces is not possible:
According to Earnshaw’s theorem (Samuel Earnshaw in 1842)
a collection of electric charges or magnetic dipoles cannot be maintained in a stable
stationary equilibrium condition by the application of static fields alone. [sku]
Earnshaw's theorem states that a collection of point charges cannot be maintained in a stable stationary
equilibrium configuration solely by the electrostatic interaction of the charges. ...
It is usually referenced to magnetic fields, but originally applied to electrostatic fields.
It applies to the classical inverse-square law forces ... [wikipedia]
• Of course one can adapt force laws. One can introduce a suitable force potential to save
phenomena. This was done by Born and Huang as Kraig et. al. [kra] mentioned:
Interatomic interaction could be expressed as a pair potential, which satisfied certain
‘reasonable’ assumptions regarding the location of its minimum and its inflection point
Obviously, 5 pair potential parameters are chosen in order to save physical phenomena...
The result of continued data fitting was a stability parameter q = p (2/a0)m.
(p, m ... potential parameters, a0 ... lattice parameter.) This parameter election is once
and for all. If potential parameters are adapted to fit the data for Pt but these parameters
don’t fit the data for Ir than the game is over, QM electron bonding theory is refuted.
• QM treats metallic lattices as virtual lattices, i.e. there are no elastic rods as in any
macroscopic lattice. Electrostatic distant forces between the lattice nodes are thought of as
only virtual bars. Of course, that mix-up of forces is not a lattice. Then for the influence of
the more distant ions or electrons Madelung gave the formula…
Metallic bonding according to QM must be explained with electrons as agents of bonding
because the standard atomic model consists of nuclei that are surrounded by extra nuclear
electrons (electron shells). So nuclei cannot be connected by material links out of the
nuclei, electron shells form an insurmountable barrier.
The electron shells of the atom model are the bed of Procrustes for molecular bonding
models...
• According to elasticity theory, minimum potential energy is the necessary condition for
equilibrium. A sufficient condition for stability of equilibrium is that the second variation
of the total potential energy should be positive for small displacements.
7
Because of the unreal lattice of QM electron-bonded molecules, applying methods of
elasticity theory must deliver faulty results.
It is necessary to emphasize that we are not concerned with small displacements. Take a
current-carrying copper cable at 100 Celsius, than we hammer it. The cable does not
explode, it does not break down. It is stable!
Take the Tacoma bridge collapse. The collapse was due to
resonant frequencies that are caused by modest wind forces.
QM stability criteria cannot include dangerous Eigen
frequencies of the lattice or of the lattice rods because there is
no genuine lattice.
• Nodal joints are not rigid;
they can undergo a permanent
plastic deformation. [hil]
• Then the unreal, virtual
lattice of QM cannot simulate
a real, solid lattice with elastic bars. Elastic bars can consist of
one or more bonds for one and the same bond length L, so buckling strength or bar
vibration Eigen frequencies (that depend on 1/L2) are expressible accordingly to bond
characteristics.
Not so the unreal bonding of QM: For one bond length there is only one force law.
One cannot ad hoc introduce different force potentials for Ir and Pt, respectively.
Excursus 2:

Here I propose an atomic model according to William Prout (1815):
Each chemical element is made out of hydrogen atoms. The so-called mass number A
represents the number of the constituting hydrogen atoms.
The crystal lattice is a genuine one with real elastic bars. The bars of the lattice consist of
hydrogen atoms.
Possible are 2 H’s in series (— —) or double H’s in series(= =) or
2 H’s parallel () or one single H (|).
Bonding is due to magnetic coupling because each hydrogen atom is a bar magnet.
Different lengths of the hydrogen atom are possible due to the elasticity of the magnetic
coupling.
Positrons are sub particles of atoms. Recall that 50% of beta decays deliver positrons.
So the proton must contain also protons.
8
The Bohr atom with extra nuclear electrons and without positrons is a failure!
Molecules are atoms conjoined in some way. Bonding in molecules is also due to magnetic
coupling. There is neither ionic nor covalent bonding (electron sharing) because there are
no extra nuclear electrons.
Bonding agents are complete hydrogen atoms and not electrons! To dock atoms it suffices
that they are nearby. Then magnetic couplings produce one or more hydrogen links.

Force virtual “lattice” versus a genuine ( = real) elastic lattice
The alternative to QM’s valence electrons and their role as bonding agents are metal
lattices that possess (magneto-) elastic bars made out of hydrogen. “Material” elastic
rods mean that they are not the QM “hydrogen bridges” that are only forces.
Here, hydrogen atoms build rods as bonds. These bonds can be single or double H-bonds.
Even two H’s in series are possible. These rods are magneto-elastic, of course.
Then one must decide which atomic cores are joined by bars.
For example, fcc structures are very stable. Bars connect atomic cores such that the
resulting basic building block is a parallel-epiped consisting of an octahedron with two
tetrahedrons attached. Octa- and tetrahedrons are Platonic bodies… See later…
Ernst Mach asked the question: Did you see an atom? Today we can see it!
But electrons in electron shells are not observables! Extra nuclear electrons can explain
some phenomena; their existence is not proven…
Free electrons as glue for metallic bonding are utterly impossible, so the alternative are
solid lattice bars even if they are not “visible” immediately so far…
There are 2 different lattice types with bcc unit cells:
 β-bcc for alkaline earth- and transition metals: diagonals and edges of the unit
cube are lattice bars
 α-bcc for alkali metals: only diagonals of the unit cube are lattice bars
Comparing the very low levels for shear, bulk and Young’s moduli for alkali metals with
the corresponding significantly greater values for transition- and alkaline earth metals, one
must conclude that alkali metals cannot possess the same lattice structure!
Alkali metals are so soft that you can cut it into pieces with a knife! Indications for a “soft”
bcc structure are also high values for Zener’s anisotropy ratios AZ for alkali metals.
Compared with hard β–bcc metals Fe, Nb, Mo, Ta, V, W, their AZ ratios are from 0.55 to
1.57. For ratios ∼ 1 metals are close to isotropy. According to this they possess greater Eand B-moduli than the soft alkali metals. The table shows some anisotropy ratios.
AZ = 2 c44 / (c11 –c12 )
AZ = 2 c44 / (c11 –c12 )
Li: 8.4
Fe: 2.4
Na: 8.3
Nb: 0.55
K: 6.7
Mo: 0.76
Ta: 1.57
V: 0.78
W: 1.01
9
The second argument for a distinction of two bcc varieties is the low melting point of the
alkali metals when we compare it with melting points of transition metals.
Melting occurs when bonds reach failure due to resonance.
That takes place when the incoming radiation frequency reaches the Eigen frequency of the
lattice rods.
Eigen frequency in turn depends inversely proportionate on L2 and the supporting mode of the rod.
L is the effective rod length that is smaller than the lattice extent because nodes are extended. The
comparatively lower melting points of the alkali’s can be explained due to a weaker
supporting mode of the rods, namely if they are simply supported, whereas the rods of bbc and fcc crystals possess the clamped- clamped support.
Therefore we distinguish two unit cells:
The comparatively lower melting points of
the alkali’s can be explained due to a
weaker supporting mode of the rods,
namely if they are simply supported,
whereas the rods of
β-bbc and fcc crystals possess the clampedclamped support.
Therefore we distinguish two unit cells:
Two different bcc unit cells:
α-bcc for alkali metals and
β-bcc for alkaline earth- and transition metals.
I α-bcc
Na
unit
cell
a/2
1 rod
= 4H’s
a=1
√3/2 = 0.866
a
10
A is the mass number of the elements. For example lithium, Li, the quantity of matter is
A = 7. According to Prout, Li is an assembly of 7 hydrogen atoms (“H’s”).
Altogether any bcc unit cell contains 2 atoms, therefore it contains 2 × A hydrogen atoms,
where A is mass number or amount of H-atoms.
Therefore the lithium unit cell contains 14 H’s.
For Li it is assumed that one rod is constructed by one hydrogen atom. The graph shows
rods of the lattice.
1: There are 8 rods inside the cell. Each bar is made out of one H. A unit cell of Li
contains 2A = 2 × 7 = 14 H-atoms. So for Li (14 – 8) = 6 H’s are left proportionally for
nodes.
2: Nodes: The red node contains 6/2 = 3 H’s.
Around a node at the cube’s vertices (shown blue) reside 8 unit cells. Therefore, the node
fraction for one unit cell is 1/8.
6/2 = 3 H’s are left proportionally for the blue nodes, so the fraction for each vertex node
of the unit cell is 3/8 H, therefore the fractions of the 8 vertex nodes of the unit cell
consume 8 × 3/8 = 3 H’s. The entire blue node contains equally 8 × 3/8 = 3 H’s.
Formula for H’s in the nodes: Mass number (2A - 8)/2 = A - 4; Li: 7 - 4 = 3
Physical properties of alkali metals indicate their lattice structure:
Due to the greater lattice constant of K, Rb and Cs, their rods must be thought of as
constructed of two hydrogen in series: ——,
whereas Li possesses a single hydrogen bond |
and Na a double bond in series = =.
Take Na-23 for example:
The number of H’s for the unit cell is 2 × 23 = 46. The bars of Na “consume” 200% more
than Li because there is a = = bonding assumed (4 H’s for each rod). Therefore 4 × 8 = 32
H’s are for the lattice bars per unit cell. 46 – 32 = 22 are left for (in sum) 2 nodes of the
unit cell. 22/2 = 11. 11 H’s are at each nodal joint.
Elasticity modulus E for α-bcc metals
The maximum elasticity modulus is in the direction <111>
(= direction of the rod) because for the other directions rods are
weakening due to missing clamped-clamped supporting modes.
11
II
β-bcc lattice type for transitions metals
+ alkaline earth metals Ba, Ra
As mentioned, rigidities of transition
metals are impossible with the weak
skeleton of the alkali metals…
Therefore, it is assumed that bars at the
edges (shown bold and red) of the unit
cube reinforce the skeleton!
Now rods can be treated as clampedclamped rods!
There are 12 edges of the unit cube,
shown red, the part concerning the unit
cell is ¼, therefore 12/4 = 3:
3 H’s are consumed for the edges of the
unit cell and 8 H’s are inside of the unit
cell. Therefore, all together the bars
consume 11 H’s.
Example V-51: the bcc unit cell contains 2A = 2 × 51 = 102 hydrogen’s.
For the number of H’s left for 2 nodes we obtain: 102 – 11 = 91. The number 91 we
cannot divide for the theoretical 2 nodes per unit cell. The nodes don’t contain the same
number of H’s. Example Fe-56, double H-bond ||: (2 × 56) – (2 × 11) = 90 H’s for 2 nodes.
One contains 45 H’s.
Data of α -bcc structures (alkali metals)
In the first row of the table below the supposed hydrogen bonding is mentioned. Bonding,
lattice spacing and other physical properties are compared. Listed are the most abundant isotopes.
α-bcc unit cell
bars: cube diagonals
single = =
bond | bond
Element
a0
100/a02
103 × 1/a04
Young’s m. E [Gpa]
d: density g/cm3
Specific heat d × C = [J/cm3K]
Melting point K
el. conductivity 107 S/m
7
Li
3.51
8.1
6.6
4.9
0.534
1.9
454
1.17
23
Na
4.29
5.4
2.9
8.93
0.97
1.16
371
2.17
2 H’s in series
——
39
K
5.33
3.52
1.2
3.5
0.86
0.65
336
1.39
85
Rb
5.56
3.2
1.0
2.4
1.53
0.56
312
0.8
133
Cs
6.14
2.6
0.7
1.7
1.87
0.45
302
0.48
Interpretation of data
E-modulus depends primarily on the number of bonds per
area: Therefore: decreasing cell parameter a0 squared
increases E.
12
Therefore E is proportional 1/a02. If the E-modulus of one single rod is not constant but
decreases with 1/a2 then E of the lattice is proportional to ∼1/a04.
This may be occur for a0 > 5 Å.
See the graph for the relation of measured E and 2000/a04 for metals Cs-Rb-K.
Due to double bonds the E modulus of Na is over two times greater than for single bonds.
Physical properties suggest that Na possesses double bonds, Li single ones:
Next table shows that melting point and bulk modulus decrease with a multiplier of 0.7 0.8 that coincides with the factor of 100/a02 for Li and Na. For slender rods
bulk modulus B is a measure of the buckling break down that is also ∼ 1/L2.
single
bond |
Element
100/a2
Bulk modulus [Gpa] B
Melting point Kelvin
Young’s modulus E [Gpa]
× multiplier =
α-bcc
7
Li
8.12
11
454
4.9
Double
bond ||
α-bcc
23
0.67
0,57
0,8
1.8 = 0.67 × 2.7
Melting occurs when rods of the lattice reach
a resonance frequency and break down. This
frequency is ∼1/L2 where L is the effective
bond length. In the table there is confirmation
that melting points of Na to Cs go down with
increasing bond lengths. Li has a short bond
length. Note that unit cell edge a0 = 3,51 [Å]
but node centers have a distance of
a*= a0(√3)/2. a* for Li is 3.04 [Å].
Na
5.43
6.3
371
8.93
100 /(a) 2
1
100 /(a)2
Cs
Rb
Cs
2MS/m
alkali melting
K/103
Rb
Li
KK
Na
Na
Li
Compare melting points of Na (371
K) with Al (934 K).
Lattice constant a0 of Na is 4.29 Å,
node-node distance is
a* = a0√2 /2 = 3.03 Å.
For Al the same magnitudes:
a0 = 4.05Å, a* = a0 √3/2 = 3.51 Å.
Due to greater rod lengths, Al should
have a lower melting point than Na.
The reverse is the case. Her, the supporting mode plays the dominant role.
Al rods are clamped-clamped rods,
whereas Na rods are supported with a
decreasing effect on Eigen
frequencies. See the explanations
under “melting”.
Similarly Li vs. Ni:
But in spite of the short Li-bond the
13
melting point of Li is very low: 454 K. Compare with fcc metal Ni that have the same a* =
3.05 [Å], but the melting point of Ni is at 1728 K.
Low melting point of Li is presumably due to supporting mode of rods.
Hypothesis: rods of Ni are clamped-clamped ones in the fcc crystal lattice whereas rods of
α-bcc Li are not so rigidly supported. (The effective bond length of Ni is a bit shorter due
to the more voluminous nodes of Ni.)
Greater thermal expansion α reduces the melting point. Graph
[] relation melting temperature/ α
Impurities: for example seeding during crystallization [hil]
Data for β -bcc metals
Cr-52
V-51
Cr/V
Fe-56
3.3
0.96
Nb
2.91
3.03
0.96
2.86
10.28
8,57
1.2
7.14
6.11
1.17
12.2 (15)
Mo-96
Nb-93
3.147
16,65
1.16
Element
W-184
Ta-181
3.1652
3.3013
19.3
Density g/cm
Mo/Nb
W/Ta
t
0.96
Ta
|
3
|
|
||
a0 [Å] unit cell
||
||
Bonding: β-bcc
||
410
190
2.16 
329
105
3.13 
294
131
4.5
6.3
0.7
4.9
7.3
0.7
4.9
8.4
2.24

0.6
211
therm. exp.1/106 K
melting K
3695
3290
1.12
2896
2750
1.05
2180
2183
1
1808
Spec heat J/cm3K
2.5
2.3
1.1
2.6
2.3
1.1
3.19
3
1.1
el.cond 106 S/m
18,2
7.6
6.6
2.8
8.7
5
0.34
0.31
0.4
0.21
0.37
1.74 
 
10
0.28
2.4 

18.2
Poisson γ
Sound vel. 10 m/s
5.22
4.15
1.25
6.66
5.12
6.63
6.09
1.09
5.93
Sound
cL/cT
longitud./tralonLongit/
Quanta cL/cT
transvers
Deby t. 300 K
1.8
2.03
1.9
2.45
1.65
2.2
2.
2.
2.
2.
2.
2.
400
240
1.5
450
275
1.6
630
3800
1.7
valency
6
5
1.2
6
5
1.2
6
5
1.2
Young E
[GPa]
3
element
a0 [Å] unit cell
100/a02;
(1000/a04)
Density g/cm
Young E
3
[GPa]
0.29
1.84
2.
Remark: for a0 [Å] > 5. → two H-bonds in series:——
==
——
Mn555555
8.91
5
Ba-138
Yb174
Ra-226
5
5.48
5.15
1.26
(1.6)
7.47
4; (1.6)
((1.6(16
3.5
(14.9(
198
13
Bonding: β-bcc
1.3
11.8
m
4.2;
(1.8)
5.5
24
therm. exp.1/106 K
22
melting K
1519
998
el.cond 106 S/m
0.6
0,3
Poisson γ
?
?
0.21
2.14
1.97
Soundvel.cL 103m/s
Lattice parameter a of Ba and Ra ∼ equal
13.2
26.3
1090
974
E depends primarily on 1/a2. But for a0 > 5. [Å]: E is not constant,
it depends on elongation, too. So E ∼ 1/a4. E for Ba and Ra ∼ equal.
E of Fe greater due to smaller unit cell and single bonds.
Great values for E, G, B for Mn due to double bonds in series
1808(Fe)/998(Ba) 0 1.82; 5(a0 of Ba)/ 2.86 (a0 of Ba) = 1.79
“wiggly” bonds —— of Ba are poor electric conductors
Sound cL/cT
1.71
1.81
1.66
Quanta cL/cT
2.
2.
2.
?
14
Remarks: Deby temp. from hyperphysics.com.
Sound velocities from [rau] cL/cT = longitudinal sound velocity/transversal sound velocity. The
ratio for longitudinal to transversal sound velocity cL/cT is treated in detail for fcc crystals. See
there.
Ultrapure barium is very hard to prepare, and therefore many properties of barium have not been
accurately measured yet. (Wikipedia)
Comparison of physical properties of β -bcc metals
Valency cannot explain essential differences of physical properties
Comparison of 52Cr with 56Fe: Cell parameter a for both, Cr and Fe, is approximately the
same, ca. a0 = 2.9 Å. But valency differs significantly: For Cr it is 6, for Fe only 3!
Coulomb force for the alleged ion forces are ≈ Q1Q2/r2, i.e. for Cr ≈ 62/32, for Fe 32/32
The Coulomb force ratio for Cr to Fe is therefore 4! But ratios for melting point and Emoduli are only 1.2 – 1.3. Therefore valency is meaningless for metallic bonding.
Melting behavior
For all metals considered, thermal expansion is low and in the same order of magnitude.
Therefore the melting points are comparable in terms of effective bond lengths.
Not surprisingly we see that there is no big difference concerning melting points if
bonding is double || or simple |.
For example W and Ta possess roughly the same melting temperature although W has
double H-bonds. Why? Melting is mainly due to resonance failures of the lattice bars.
At the melting point the resonance frequency is achieved for the bonds and the number of
bonds does not matter – all bonds break down! (Exception: If a metal possesses rods with
different lengths then longer rods collapse first. There are two melting points – depending
on rod length.)
Compare W with Mo: melting point of W is higher because its bond length is shorter due
to greater atomic dimensions.
The same behaviors show Ta-181 and Nb-93. According to current theory melting point of
Ta should be roughly twice of Nb because the mass ratio is twice. This not the case. The
melting temperature ratio is Ta/Nb = 3290/2750 = 1.19! Predominant is not mass but
effective bond length of the lattice. For both, T and Nb, the lattice parameter is the same:
3.3 Å. Because the Ta-nodes are greater than the Nb nodes, the effective bond length of Ta
is shorter. Therefore the resonant frequency of Ta bonds is higher; therefore the melting
temperature must be higher…
The rationale of the temperature-pressure curve
The graph [br] shows temperaturepressure curves for various metals. One
might assume that rising pressures
cause lower melting temperatures. The
reverse is the case. Why?
Take Al for example. Lattice constant
is large, respectively: 4.05Å. Melting
occurs due to resonance, resonance
frequency depends on 1/L2 where L is
the effective bond length. But pressure
shortens and thickens the lattice rod a
15
bit, makes it stiffer. So the melting temperature increases!
Now compare W and Fe: melting point of W is remarkable higher than for Fe. Among
metals, W possesses the highest melting point: 3695 K.
W possesses double bond rods, Fe also.
The cause for higher melting temperatures of W is the difference regarding effective bond
lengths! Lattice constants for W-184 and Fe-56 are 3,17 Å and 2.86 Å, respectively. But
because of the greater nodes of W-184, the effective bond length of W is reduced; it is
smaller than the effective bond length of Fe. Therefore W melts at higher temperatures.
The graph shows that an increase of pressure from 60 to 100 GPa does not further increase
the melting point. Obviously more pressure cannot alter remarkably the effective bond
length. A further increase of the pressure causes the buckling of the rods. The result is
melting.
It is interesting to compare W-184 with Ta-181. In order to explain different and equal
physical properties one must assume that Ta has single bonds whereas W has double
bonds. Lattice constants don’t differ remarkably. Double bonds of W cause a 2.2-fold
amount of Young’s modulus E and a 2.5-fold better electrical conduction. Melting points
are almost equal because the resonance frequency is equal for bonds of the same effective
length. Both bonds of the double bonds break down due to one and the same resonance
frequency.
E- moduli indicate the nature of bonding. In comparison with V, Nb, Ta the elements Cr,
Mo, W have significantly greater moduli.
Empirical data show that these moduli depend on the dimensions of the lattice constants
and of the character of bonding, i.e. if the bond is a single or double H.
The high values of E-, G- and B- moduli of Ce, Mo, W are due to a strong bonding:
The bonds consist of two H’s parallel.
The other metals (V, Nb, and Ta) possess E-, G- and B- moduli that are one order of
magnitude lower. This indicates that bonding is fragile; it is assumed that one single H
forms the bond.
As shown in the table, shorter bonds are stronger bonds. Reduced bond lengths due to
nodal mass contents are unknown and may cause deviations from the rules mentioned.
Young’s modulus E
… dependence on unit cell structure: β-bbc = rigid; α-bbc = soft
There is evidence that E depends roughly on 1/a02, when lattice constants a0 are roughly in
the same range. Especially for widened bonds, dependence is roughly on 1/a04, see above.
Take for instance E modulus for Rb and Nb. Because mass numbers A show no big
difference, the effective bar length is roughly proportional to the lattice constant a0.
But the crucial role for the modulus E plays the lattice structure.
Lattice constant
a0
4
[Å];
(1/a
)
Young’s modulus E
[GPa]
Bulk modulus
B
Rb, A = 85,5
Nb, A = 93
Nb / Rb
5.56;
3.3; β- bbc
0.6
2.4
105
43,75
2.5
170
68
α-bbc
[GPa]
In comparison with Rb, Nb possesses the 44-fold modulus E because its lattice constant is
16
roughly only half as long. Then the crucial point is that Rb has a α-bbc lattice whereas Nb
is β-bcc. α-bbc crystal structure shows a weaker lattice cell than β-bbc, see above the
figures.
Nb possesses one strong and short bond whereas the weak bonding of Rb consists of two
H-bonds in series. Nb possesses the 68- fold B-modulus because the α-bbc lattice of Rb
can easily be compressed.
More examples: Ratio of E moduli W/Mo = 345/329 = 1.05. The corresponding ratio
of unit cell parameter a0 is 1.006. Compare E ratio W/Ta=345/186 = 1.85. The multiplier is
mainly due to the 2:1 ratio of bonding | | / |. For Mo/Nb the E-ratio = 2.6, for Cr/V = 2.2.
(There are many different moduli for E measured!)
Conclusion
Electron bonding model of QM cannot explain remarkable differences of G, E, and B
modulus for almost equal unit cells and atomic weight.
Valency is meaningless for metallic bonding.
———————————————————————————————————
fcc crystal system
There are 12 octahedron bars (shown blue) and 24 face
bars that are shared 50% between two incident cells.
12 + 24/2 = 24. 24 H’s are consumed for the bars.
How many H’s contains a node? Altogether the unit cell
contain 4 atoms, therefore
4 A is the amount of H’s in the unit cell, where A is the
mass number.
Therefore 4 A are shared between bars and nodes.
Therefore (4A - 24) is the number of H’s that 4 nodes
contain. (8 nodes in the vertices of the cube consume
8/8 = 1 and 6 octahedron nodes consume 6/2 = 3 parts of
the remaining H’s.)
Therefore the equation holds:
(4A -24)/4 = (A – 6) = number of H’s in the nodes.
Example: 63Cu: 63 – 6 = 57.
If the allocations of the atoms mark the nodes of a real lattice then one must decide which
nodes are connected by bars. So wanted are stable building blocks of metals.
Lattice building bricks must be space filling. Cubes are space filling.
Alternate building brick: The minimum unit cell is an octhedron plus 2 tetrahedrons, see
figure. The maximum elasticity modulus E is in the direction <111> .
(Conjecture: The hydrogen atom is thought of as to consist of electrons and positrons. The
electron is electromagnetically connected to the proton. The proton is not an elementary
particle but consists of sub particles.)
17
Bonding
Due to its physical properties, bonds are thought of either as
single bond |;
as double bond parallel ||;
as single bond in series ——;
as double bond in series = =
With double bonds that consist of two hydrogen’s parallel || → metals with maximum
stability. Fcc metals possess high heat capacities, are good conductors, and are rigid.
Verification of the claim that an fcc unit cell contains 4 atoms:
The Rh-mole has 103 grams. The total molecular weight of the atoms in a unit cell is
4 x (103) = 412. The weight of the 4 atoms in the Rh unit cell is therefore
412/NA = 68,426 x 10 -23 g (NA = Avogadro number)
The density of Rh is weight/volume. The volume of the unit cell is lattice constant a cubed.
When a0 = 3,8034 10-8 cm , a03 =55.02 x 10-23. Density: 68.426 10-23/55,02 = 12,44 g/cm3,
in good agreement with the reported values. Therefore the claim that the fcc unit cell contains the mass of 4
atoms of the element is verified. The same can be done for a bcc unit cell.
18
Diamond: highest melting point
Mass number A is for the most abundant isotope. 6C is diamond.
There are different values for physical properties listed.
——
Bonds
||
|
element
C
Ir
Rh
Ni
Pt
Cu
Pd
Ag
Au
Al
Ca
Sr
Mass # A
6
193 103 60
195 63
106 107 197 27
40
88
lattice a0 [Å]
3.57 3.84 3.8
3.52 3.92 3.62 3.89 4.09 4.08 4.05 5.59 6.08
Young GPa
1141 525 384 200 169 130 121 121 78
72
20
15.7
Therm exp.1/106 K
6.4
8.2
13
8.8
16.5 12
18.9 14.2 23
22.3 22.5
Melting K ÷103 3.82 2.74 2.24 1.73 2.04 1.36 1.83 1.23 1.34 0.93 1.11 1.05
valency
4
6
6
2
6
2
4
2
5
3
2
2
C = diamond, Lattice constant a0= 3.57 Å, nearest neighbor distance 1.55 Å!
Rhodium E- measurements show 373-384 GPa! Source: J. Merker et.al. :Platinum Metals Rev. 2001,
Pb
208
4.95
16
29
0.6
4
45,
(2)
Comparisons:
Melting points:
Diamond’s highest melting point due to smallest neighbor distance.
Regarding E moduli of Ir-193 (525 GPa) and Pb-208 (16 GPa), the huge difference is due
to bonding: Ir has strong and short double H-bonds, whereas the bonds of Pb are two H’s
in series that are the cause for a relatively huge bond length.
Comparing E-, G- and B- moduli of Ir-193 with Pt-195, respectively, significant
differences are not understandable in terms of current theory: Ir and Pt, both possess
roughly equal bond lengths and identical valency, so they should have also equal physical
properties!
The hydrogen-bonded metal lattice (where bonds are elastic (real) H-bridges) can explain
the differences regarding E- moduli:
Double H-bonds || (for Ir) are the cause for greater E modulus. Single H-bonds | (for Pt)
cause lower E-modulus. Shorter bond lengths cause stronger bonds.
The ratio of mass number A for Au to Al is 197/27 = 7.3; the nodes of Au contain
197 - 6 = 191 hydrogen atoms whereas the Al nodes contain only 27 - 6 = 21 hydrogen’s.
Therefore, the bond length of the Au lattice is more reduced than the bond length of the Al
lattice. Because of its comparatively more reduced bond length; Au possesses a greater
buckling resistance! If L of Al is set 1.00 then L of Au is 0.577 because 1/L2 =1/0.5772 = 3.
(Supporting mode unknown…)
Also Vickers hardness can serve as a measure of bond strengths. The ratio Au/Al =
216/167 MPa = 1.3
Shear moduli G and elastic moduli E of Au and Al don’t differ significantly. Why? These
moduli depend on number of bonds per area and the areas of the unit cells don’t differ:
lattice constant of Au is 4.08 for Au and 4.05 Å for Al!
Citation below discovers that frequencies of gold “lying appreciably higher than those
“scaled” from copper and silver…”
Authors attempt to explain in terms of electromagnetic forces of the orthodox model.
The rod-lattice model of metals delivers a simple explanation: compared to Cu and Ag, the
effective rod length of Au lattice is smaller than for Ag and Cu. Therefore the Au lattice is
more stiff and therefore the Eigen frequencies are higher! According to the presented
model here, Au nodes contain (197- 6 )= 191 hydrogen atoms, Ag nodes contain (107- 6)
or (109- 6) hydrogen atoms. So for Au the rod between the nodes is smaller than for Ag
and frequencies are higher…
19
*
Phys. Rev. B 8, 3493–3499 (1973): Lattice Dynamics of Gold J. W. Lynn , H. G. Smith, and R.
M. Nicklow
The complete phonon dispersion relations for gold in the high-symmetry directions have been
measured at room temperature by the coherent inelastic scattering of neutrons. It is found that the
forces in gold are not homologous with the other noble metals, the frequencies of gold lying
appreciably higher than those "scaled" from copper and silver. An analysis of the data in terms of
different force-constant models reveals that a general tensor force is required for the first-neighbor
interaction, whereas for neighbors beyond the first either general tensor or axially symmetric forces
give an excellent fit to the data. The axially symmetric model alone does not adequately describe
the data even when forces extending to ninth-nearest neighbors are included in the fit. In addition,
simple screened-pseudopoential models were fit to the data and these results also indicate the
need for the first-neighbor interaction to be general. Frequency distribution functions and related
thermodynamic quantities were calculated from the various force-constant models. The Debye
temperature ΘC versus temperature curves obtained show an anomaly at low temperatures
consistent with the ΘC(T) obtained from specific-heat measurements. The relation between this
anomaly and the character of the dispersion curves is given.
© 1973 The American Physical Society
http://link.aps.org/doi/10.1103/PhysRevB.8.3493
An open question of Ca is it’s bonding: electrical conductivity indicates that Ca belongs to
the best conductors like Ag, Au, Cu and Al that are fcc with a single bond. On the other
hand Ca possesses comparatively low physical properties like E and G moduli that are due
to the comparatively big lattice constant of Ca. Ca bonds could consist of two H’s in series.
So Ca would be also a good electrical conductor: The vibrating bonds can be polarized.
Ca-40 is paramagnetic. Strong supporting mode of the rods could explain the relatively
high melting temperature of Ca. Recall that over 721 Kelvin, Ca possesses a bcc structure.
Because of the smaller shear moduli and great lattice spacing’s, for Pb, Ca, and Sr it is
assumed that they possess 2 H’s in series as bonds.
Not listed is Ge-73. Values are: a0 = 5.66 [Å], d = 5.3 g/cm3, E = 80 GPa, melting at 1211
K, sound 4630 m/s. Because of the relatively high E modulus the bonds of Ge are probably
double bonds in series = =.
Bonds
element
C
||
Ir
Rh
|
Ni
Pt
Cu
Pd
Ag
Au
Al
——
Ca
Sr
Pb
Mass # A
6
193
103
60
195
63
106
107
197
27
40
88
208
lattice a [Å]
3.57
3.84
3.8
3.52
3.92
3.62
3.89
4.09
4.08
4.05
5.59
6.08
4.95
E Young GPa
1141
524
373*
200
169
130
121
121
78
72
20
15.7
16
Poisson γ
0.15
0.26
0.26
0.31
0.38
0.34
0.39
0.37
0.44
0.35
0.31
0.28
0.44
6.45
3.62
c
cL/cT
2.
1.75
1.75
2.01
2.28
2.03
2.35
2.35
3.07
2.08
1.91
1.8
3.06
Quanta ratios cL/cT
2.
2.
2.0
2.0
2.0
2.0
2.
2.
2.0
2..0
2.
2.
2.0
Sound
Legend: cL/cT = longitudinal sound velocity/transversal sound velocity
* E 373-384 GPa according to Platinum Rev. 2001, 45, (2)
Comparisons of sound velocities due to crystal structures
C = diamond, Lattice constant a = 3.57 Å, nearest neighbor distance 1.55 Å.
Sound longitudinal:18350; transversal 9200 m/s, cL/cT = 18350/9200 = 2!
Nickel: cL/cT = 6040/3000 = 2.01, expected 2.0; nickel unmagnetized: cL/cT = 5480/2990 =1.83!
Source: http://www.rfcafe.com/references/general/velocity-sound-media.htm
20
Poisson’s ratio can be written a) in terms of B and G moduli, or b) in terms of longitudinal
and transversal speed of sound:
a) γ = [3(B/G–2)]/[6(B/G+2)];
b) γ = [1/2(cL/cT )2-1]/[(cL/cT )2 - 1]
For Ni, Cu, Al the cL/cT ratio for sound is about 2. Listed Poisson ratios are 0.31-0.35.
For a cL/cT ratio of exactly 2, Poisson’s ratio is 0.33. When we assume that the true value
is 2 , then it follows that cL/cT ratios are quantified!
But Ir compared to Rh have equal lattice constant’s, show equal Poisson and cL/cT ratios
despite different atomic numbers (193 v. 103). Shorter effective bond length of Ir is
responsible a higher E-modulus value.
If, for instance, the longitudinal sound wave velocity for Cu is two times greater than the
transversal sound velocity, then (for the same frequency) the longitudinal wavelength is
two times greater than for the transverse wave…
Conjecture: There are quantized Eigen frequency ratios cL/cT (quanta).
It is striking that for C this ratio is exactly 2.
Now we compare longitudinal and transversal waves for cubic unit cell metals in the
<111> direction. According to [www.pa.uky.edu/~kwng/phy525/lec/lecture_3.pdf] the ratio of
these velocities
vL / vT = √{(c11+2c12+4c44) / (c11-c12 + c11)}
Calculated values for fcc metals are: Al (2.14); Cu (2); Ni (2.16);Ag (2.55); Au(3.1).
For bcc metals: W (1.8): Ta (2.2); Mo (1.78); Fe (2.15). Compared with measured sound
longitudinal and transversal velocity ratios, they are probably more adequate. All these
values are with respect to polycrystalline metals.
For a regular array of crystals we assume the following ratios:
Al (2.0); Cu (2.0); Ni (2.0); Ag (2.); Au (2.0 – not 3.0 ?).
For bcc metals: W (2.): Ta (2.0); Mo (2.); Fe (2.). The ratio for hcp metals is also 2.0.
hcp
close packed hexagonal metals
Remark to the figures: distance d (node-node centre) is of the inclined rod of the lattice
model! The hcp unit cell contains the amount of 6 atoms, therefore 6 × A H’s, where A is
mass number of the element. Rods of the unit cell consume 30 H’s. Nodes of the lattice
contain (A - 5) H’s. Without details, see the method for fcc.
Co is remarkable because the structural ratio c/a0 = 1.633. Therefore d = a0.
In the 5th row are node centre - node centre distances. Effective rod lengths L depend on
unknown node dimensions. It is remarkable that lattice distances a0 and d roughly
coincide for most hcp metals!
21
 hcp
Be
Co
α
Mass #
9
59
|
1,85
2.27/
3.6
2.22
19.4
287
1560
2,50
0.032?
1.98
2.
g/cm
a0/c
3
d =√(a2/3+
c2/4) 02
100/a
E GPa
Melting K
el.c.107 S/m
Poisson γ
Sound
cQuanta
L/cT
cL/cT
Zr
Ru
Os
Mg
Ti
Sc
Y
90, 94
102,10
4
192,19
0
24, 26
48, 46
45
89
|
|
||
||
——
—
—
—
8,9
2.5/
4.1
2.5
16
209
1768
1,7
0.31
1.91
2
6,5
3.2/
5.1
3.18
9.6
68
2128
1.7
0.34
2.03
2.
12,4
2.7/
4.3
2.65
13.6
447
2607
1,4
0.3
1.87
2
22,6
2.73/
4.3
2.68
13.4
558
3306
1,2
0.25
1.73
2.
1,74
3.2/
5.2
3.19
9.8
45
923
2,24
0.29
1.83
2.
4,51
2.95/
4.7
2.91
11.5
116
1941
0.24
0.32
1.95
2.
2,99
3.3/
5.3
3.25
9.2
74
1814
0.15
0.28
1.81
2.
4,47
3.64/
5.7
3.55
7.5
64
1799
0.18
0.24
1.71
2.
Legend: a0 and c are (conventional) unit cell dimensions. (The ideal c/a0 ratio is 1.633).
The c/a0 ratio of these unit cells is in a range of 1.57 – 1,62. The ideal ratio is 1.63, where
a0 ≈ d. Special case Be: Take account of the small mass number of Be (A = 9), for which
reason a considerable part of the specific heat per volume is probably due to oscillations of
the lattice nodes!
Zn and Cd have c/a0 ratios 1,89 and 1.86, respectively.
Author’s conjecture: High rigidity of the lattices of Os and Ru is due to a double
hydrogen bond ||, whereas the other metals possess a single H-bond |.
Remarkable is a comparison of Ru with Os: their cell dimensions are the same, but the
Osmium lattice is more rigid. Why? Because the lattice nodes of Os are greater, the bonds
of Os are shorter and therefore stronger. Big elastic moduli E of Ru and Os are due to
double bonds.
Compare Mg to Sc: Effective bond length of Sc is smaller because the Sc node is greater
than the Mg node. Therefore the melting point of Sc is higher.
In all cases the bonding varieties play no role, only the effective lengths of the rod
determine resonance frequencies and melting points.
Mg and Zn show the lowest melting points. It is possible that the isotope mixture of these
metals changes physical properties!
Remarkable is Co:
Cobalt occurs as two crystallographic structures: hcp and fcc. The ideal transition
temperature between the hcp and fcc structures is 450 °C, but in practice, the energy
difference is so small that random intergrowth of the two is common.[wikipedia-cobalt]
Remarkable is Beryllium: It has the smallest cell that is very rigid. Comparison of ratios
longitudinal to transversal sound velocities cL/cT = 2.0
Take for example Os, Y, Lu, γ = 0.24 - 0.26; cL/cT = 1.75, for a regular array of single
crystals cL/cT = 2.0 (thesis).
Ti: cL/cT = 6100/3050 = 2.0!
22
dhcp structure = double hexagonal close packed structures
wickipedia Periodic table (crystal structure) introduced the term dhcp structure instead
of simple hexagonal structure. Next table shows metals with double hexagonal packed
structures (graph:wiki).
Double
hexagonal
Zn-64, 66
Cd-144
bonds
——
c/a0
E GPa
Melting K
La-139
Pr-141
Y-89
——
——
——
——
1.86
1.89
12.14/3.77=3.22
11.8/3.67= 3.2
5.73/3.64 =1.7
108
693
50
594
36.6
1193
37.3
1204
64
1799
dhcp metals are:
Zn, Cd. La, Pr, Nd, Pm Am, Cm, Bk and Cf.
The table shows that for the pair La and Pr mass # A, density and cell
dimensions c/a0 are roughly the same. So E and melting temperature are
also roughly the same.
Melting temperatures of La and Pr are high despite long c-rods, why?
There are two melting points: melting begins with the failure of the long
rods (lattice constant c). Only then the short rods (a0) fail. This is the
measured melting point!
The values for a0 are 3.67 Å and 3.64 Å for Pr and Y, respectively.
Melting point for Y is 1799 , and for Pr 1799 K due to the short a0!
Open questions: Effective bond length of the long bonds c ? Supporting modes of the
exorbitant long c-rods (∼12Å)?
Excursus: melting, buckling
Beam: transversal vibration, Eigen
frequencies
Radial Eigen frequency: ω = a (nπ/L)2,
a = EI/ρA, A… area, E…Young modulus,
I area moment of inertia,
n = 1, 2, 3,…ω = EI/ρA (nπ/L)2
Circular beam, radius r, I = (r4)π/4 ; I/A = r2/4.
a = E r2/4ρ, ω = (nrπ/L)2 (E/4ρ),
i.e. the shorter and thicker the beam the greater
the Eigen frequencies.
23
Note quantization of kn factors of Eigen frequencies in the table:
support
k1 : k2
: k3
Clamped-clamped
1
: 2.75 : 2
Pinned-simple supp.
1
: 4
: 2.25
Free-free
1
: 2.75 :
2
Clamped.simple supp.
1
: 3.25 : ∼ 2
Clamped- free
1
: 6.25 : ∼ 2.75
The greater the Eigen frequencies the greater the specific heat per volume. (See the paper
on specific heat).
Example Na-23: double H’s in series (= =); nearest neighbour distance = 3.72Å;
volumetric specific heat: J/cm3K = 1.16. Rh-107: 2 H’s parallel, nearest neighbour distance
= 2,69 Å, effective rod length shorter due to big nodes; spec. Heat/vol.:J/cm3K = 3
(Source: A.V. Metrikine, dynamics, slender structures…textbook, delft technical
university.http://www.mechanics.citg.tudelft.nl/~andrei_m/)
Melting
Empirical data suggest that melting points of metals depend
on 1/L2, where L is the effective bond length. Why?
Eigen frequencies of bars depend on 1/L2. When the
oscillations of the lattice bars achieve resonant frequencies,
failure occurs. Again the comparison Na vs. Rh. Rh possesses
the smaller effective bond length. Melting temperature of Rh
is therefore significantly higher than for Na: 2240 K vs. 371
K! Note that Na is bcc, Rh is fcc, supporting modes can be
different.
The QM bonding model cannot explain the melting point due to critical velocities of
electrons and ions...
Melting explained due to resonant lattice failure is superior to QM explanation.
When you heat a solid below its melting point, the heat which the solid absorbs raises its
temperature. When the solid reaches its melting point, it continues to absorb heat but it
stays at the same temperature until it has melted completely after it has all liquefied, the
liquid will again rise in temperature as it absorbs heat. [hol]
So during melting it is not necessary to increase the temperature! Reason: At the melting
temperature the lattice breaks down because a resonance frequency was achieved.
No more energy is needed to fulfill the job.
In terms of QM, increasing the temperature above the melting point should increase the
movements of the carriers of heat, namely the alleged free electrons of the metal. This
increase of temperature was not observed!
Melting criteria according to wikipedia:
... The Lindemann criterion states that melting occurs because of vibrational instability,
e.g. crystals melt when the average amplitude of thermal vibrations of atoms is relatively
high compared with interatomic distances...
24
The Born criterion is based on rigidity catastrophe caused by the vanishing elastic shear
modulus, e.g. when the crystal no longer has sufficient rigidity to mechanically withstand
load.
Comment: Lindemann is right, vibrational instability is the cause for melting but instability
concerns the lattice bars that don’t exist for Lindemann and Born!
Melting of alloys
Wikipedia: Unlike pure metals, most alloys do not have a single melting point, but a melting range
in which the material is a mixture of solid and liquid phases. The temperature at which melting
begins is called the solidus, and the temperature when melting is just complete is called the
liquidus. However, for most alloys there is a particular proportion of constituents (in rare cases
two)—the eutectic mixture—which gives the alloy a unique melting point.
Explanation: An alloy of two metals possesses crystals with different bond lengths due to
different magnitudes of the lattice nodes. Melting begins at the longer bonds where the
resonance frequency is lower, and then at higher temperatures melting of the shorter bonds
follows.
Relation of melting magnitudes versus 1/L2
Regarding Cu, Au, the melting point is roughly equal, but 1/L2 for Au is greater.
But thermal expansion of Cu is greater than for Au, therefore the Cu bond length expands
more and near the melting point bond lengths of both Cu and Au are ∼ equal, so the boiling
points are also ∼ equal. The table shows more examples to explain in this manner.
Regarding Ca and Sr, both have the same thermal expansion, therefore they are
comparable with respect to melting. Because melting is due to a resonance catastrophe and
resonance frequencies depend on 1/L2, Ca has a higher melting point because its bond
lengths are smaller.
Melting points and shear moduli don’t correlate because they are measuring other
quantities… Melting begins when bullets of hot gases hit the bars and destroy them, or
when the bars crash due to resonance frequencies. A bar in a lattice can crash also when its
fixing in the nodes is too weak…
Heat capacities predominantly depend on length. Lengths determine Eigen frequencies
with its overtones. Melting points should correlate with 1/L2 but with increasing
temperatures the length L at 20 C˚ increases too. But the thermal expansion coefficient
itself is a function of temperature and metals differ with respect to thermal expansion.
Example Na vs. Ba: Melting points differ ∼ 1:7, but 1/L2 differs not with this order of
magnitude. Explanation: Thermal expansion of Na is greater than thermal expansion
of Ba (∼ 4:1). Therefore, Na bonds expand more than Ba bonds do. Consequence: Na
possesses a very low melting point.
For β-bcc metals there are:
1: node centres that have a distance of a0 and
2: node centres that have a distance of a*= a0 (√3)/2.
Melting begins at the longer bonds but is finished only after the smaller bonds break down.
For β-bcc crystal structures there are strictly speaking two different melting points. But
technical failure begins with the collapses of the longer bonds…
In the table for bcc metals the dependence of melting and bond lengths is shown. Melting
points decrease from W to Fe. The first 4 metals compared are W, Ta, Mo and Nb, all are
β-bcc metals. Ta and Nb possess the same lattice constant a0, but different melting points.
As explained, the effective bond length of Ta is smaller as the bond length of Nb because
the nodes of Ta are greater. So the decrease of melting point is explainable
25
Basic structural building block of fcc structure is an octahedron with two tetrahedrons
attached, edges possess the same length a*= a0 (√2)/2. In the table there are the relevant
values a* computed.
bcc and fcc metals cannot be ordered with decreasing melting temperatures. So bcc and fcc
metals seems to be different with respect to melting. Compare W-184 with Ir-192. Bond
lengths are roughly the same but melting point of W is about 1000 K higher…
One cause is that bcc crystals possess two bonds with different lengths … Then Ir and W
are isotope mixes.
Compare in the table below decreasing melting points of Cr and Fe. But Fe bond lengths
are roughly the same.
Why? One explanation may be the isotope mixes of Fe and Cr, both have 4 stable isotopes.
Ta181
Mo-96
Nb-93
V-51
Cr-52
Fe-56
Crystal structure
W184
β-bcc
Lattice a0 [Å]
3.17
3.3
3.15
3.3
3.03
2.91
2.87
fcc: a*= a0× 0.5√3
2.75
2.86
2.73
2.86
2.62
2.52
2.49
Melting K
3695
3290
2896
2750
2183
2180
1811
Lattice a [Å]
Ir-192
3.84
Rh-103
3.8
Pt-195
3.92
Pd-106
3.98
Ni
3.52
fcc: a*= a0× 0.5√2
2.71
2.69
2.77
2.81
2.49
Melting K
2739
2237
2041
1828
1730
Buckling strength of crystal lattice rods
Leonhard Euler derived a formula for the critical buckling load:
F = π2EI/(KL)2 where F = elasticity modulus, I = moment of inertia, L length of column,
K = a factor that depends on the mode of supporting. For fixed ends: K = ½.
Critical longitudinal buckling load modes (n = 1, 2, 3): N cr, n = EI (n π/L)2
See for boundary conditions: University of Ljubljana. esdep course on stability. http://www.fgg.unilj.si/kmk/esdep/master/wg06/toc.htm
Buckling depends on 1/L2
Conclusion: Beam Eigen frequencies as well as critical buckling loads depend on 1/L2
The melting point of a metal is achieved when resonant frequencies and /or buckling loads
destroy the crystal bond. This may be due to temperature, pressure or “bombardment”.
If empirical data show a correspondence between boiling temperatures and the inverse
square of bond length the bonding model due to central electrostatic distant forces between
free floating electrons and ionic atomic cores is no more applicable.
Bonding must be considered as a problem of structural mechanics, where instability of
equilibrium can be due to buckling and resonant oscillatory modes.
The elastic bonds are thought of as hydrogen atoms. Such a crystal is H-bonded and not
electron bonded as QM ascertains.
26
20
Table IX: Predicted melting temperatures.
Material
Tm (K)
Theory Experiment
Li (bcc)
620
454a
Ca
650
1124a
Al
1150
933a
Mo
3320
2890a
Ir
4230
2683a
FeAl
1630
1583b
CoAl
2040
1921b
NiAl
1700
1911b
Ni3 Al
1890
1668b
RuAl
2380
2193b
RuZr
2770
2373b
SbY
1590
2583b
TiAl
1370
1746b
NbIr
2830
2113b
a Experimental data from Kittel (1986)
b Experimental data from Massalski et al. (1986)
← Predictions of shear modulus and
melting points according
to current QM electron bond theory
Many first principles calculations of
the electronic structure and total
energy of solids have
been carried out...[meh]
But the results are meager and refute
QM metallic bonding!
Regarding predictions of melting
temperatures according to the rough
convergence between the average
elastic constants (C11 + C22 + C33)/3
the following table shows remarkable
discrepancies between theory and
experiment:
Comment of Mehl et.al:
The high melting temperature
materials such as Mo, Ir, and Nb, Ir
fall well outside of the 300 K error
Summarybars of the prediction.
Remarkable is Ir: Its theoretical melting temperature is about 60% over the experimental
have shown that
it is possible to use the formalism of Density Functional Theory and
value!
Local Density Approximation
to calculate
thehigh
equation
of state
and elastic
constants
of that Eigen
Above the explanation
of the
melting
temperature
of Mo
and Ir was
2
2
ple metals and ordered
binary
intermetallic
alloys.
In
most
cases
the
equilibrium
lattice
frequencies of lattice bars depend on 1/L : ω = EI/ρA (nπ/L)
stants are underestimated by 1-2%, which is typical of other LDA calculations. A small
tion of this error
represents
the error
neglecting
zero-point
motion
thermal
Mo
and Ir possess
shortindouble
H-bonds;
therefore
theyand
possess
alsoexhigh Eigen
sion, the remaining
error
is
from
the
LDA.
We
have
also
shown
that
we
can
successfully
frequencies with the consequence of high melting temperature. For current theory of
dict the elastic constants
of these
to within
10%
experiment,
electrostatic
forcesmaterials,
as bonds,typically
these forces
depend
onofdistance
only.with
aximum error of about 20%. These elastic constants can be used to estimate the shear
dulus in polycrystalline
materials,
where
again successfully
compared
our results
with
Above we
explained
whywecurrent
theory couldn’t
reasonably
predict
shear modulus G. In
eriment. Finally,
we
showed
that
the
elastic
constants
are
roughly
correlated
with
table VIII of the work cited there are the values for G by experimentthe
and by distant forces
ting temperature
of the solid.
equilibrium
theory for Mo, Ir, Ca, Al and Li.
What is the future
first-principles
calculations?
addition
tounderlying
“predicting”
the of the model.
Theseofpredictions
are good
but cannot In
prove
true the
ontology
ting temperature
using
(39)
and
(40),
we
have
also
looked
at
the
barriers
which
prevent
Above mentioned anomalies regarding shear modulus remain unsettled!
lattice from “hopping” between various phases with the same structure but different
ntations. There is some indication that this “magic strain barrier height” is correlated
he melting temperature (Mehl and Boyer, 1991). Simple models of defects have also
n studied. In particular, first principles calculations have been used to determine the
material
G theory G exp.
G exp. new
ancy formation energy in Aluminum (Dentenerr and Soler, 1991; Mehl and Klein, 1991;
Mo (bcc)
119
130
126
edek et al., 1992). This number is important for studying defects in solids and the
Ir (fcc)
226
221
210
ion of dislocations, which can be pinned by defects. The first principles calculations
Ca (fcc)
8.1
8.9
7.4
uired to do further work in this area are rather complex and computer intensive, but
Al (fcc)
28 which can26be handled
26by the current generation of
within the range
of calculations
Li (bcc)
3.4
3.9
4.3
27
Crystal structure determines stiffness
metal
Na
(α-bcc)
Ca
(fcc)
Bond length [Å]
G shear modulus GPa
G value due to:
3.72
3.95
3.3
7.4
greater G value of
Ca due to fcc
structure
V, Fe
(β-bcc)
Cr
Ir
(β(fcc)
bcc)
2.73 vs. 2.57 2.7
2.71
47 vs. 64
115
210
V, Ni: single H bond |
Fe, Cr, Ir: double H bond ||
Ni
(fcc)
2.48
76
Bond lengths: a0√2/2 for fcc; 0.9a0 for β-bcc ; a√3/2 for α-bcc.
Atttention: the shear modulus of Fe is that of ductile or malleable Fe, not of structural
steel!
Iron, Ductile
Structural
Steel
(GPa)
63 - 66
79.3
Zener’s anisotropy ratio AZ
Preliminary remaks:
Real macrosopic pieces of metals are polycristalline
aggregates. This means a conglomerate of metal grains
that possess different directons of physical properties
like elasticity, electrical conduction.
Single crystals don’t exist. To do experiments there are
agglomerates of single crytals necessary, at least several
milimeters large. Obviously, for Al such a probe was
available and the elasticity moduli were directly
measurable. See below.
Of course, due to the fcc structure, Al is anisotropic.
The polycrstalline Al is roughly isotropic.
To determine anisotropic E moduli of Al the experiment with single crystals is meaningful.
Alternatively one can calcualte the elasticity moduli with elasticity factors Cij:
Due to symmetry, for cubic cristals remain 3 independent elasticity factors
C: C11, C44, and C12 .
28
Zener proposed for the anisotropy ratio AZ = 2C44/(C11 - C12)
Tables for Zener’s anisotropy ratio AZ and E111/E100 ratio [rol]
bcc
V
Nb
Fe
Ta
W
Zener’s AZ
0.78
0.55
2.41
1.57
1.01
1.5
1.01
E111/E100
2.15
←
→
[webclass]: W: E111= E100 = 385 GPa. Exp. E = 411 GPa!
E111/E100
0.8??
0.57
fcc
Al
Ni
Cu
Pt
Ag
Pb
Pb
δ-Pu
Zener’s AZ
1.22
2.45
3.21
1.59
2.92
4
4
7
E111/E100
1.19
2.18
2.87
Compare V to Fe:
bcc
V-51
Fe-56
a0 [Å]
3.
2.9
Zener’s A
0.78
2.41
E111/E100
0.8
2.15
E111 calc
E100
E [GPa]
measured
117
146
270
125
130
211
E [GPa] of polycristalline Fe due
to smaller unit cell
area and thicker
rods...
Ranking of hardest metals, E-moduli [GPa]:
0: hcp-Os (560?)
1: fcc-Ir (528).
2: hcp-Re (463). 3: hcp-Ru (447).
4: β-bcc-W (411). 5: β-bcc-Mo (329)
Why first places for Os and Ir?
1: hcp and fcc structures, the most stiff structures.
2: Effective bond length are mall, so rods are more compact and E-moduli of rods increase.
Effective bond lengths are comparatively small due to great mass content number in the
nodes.
Same considerations for first place of W among bcc metals...
———————————————————————————————————
Anisotropy in Aluminium
http://aluminium.matter.org.uk
Graph showing effect of single crystal orientation
on the Young modulus of Al.
As the cube axis is rotated from 0°/90° to 45°
to
the
tensile
axis,
the modulus increases from 63 to 72 GPa, or
about 15%.72/63=1.142857143=8/7
Experimental E111/E100 ratio for Al is 1,1428,
theory delivers 1.19.
29
δ-plutonium (fcc)
According to current theory for δ-plutonium (fcc) there is a huge Zener anisotropy
AZ = 7! So the interatomic bonding is an open question for the theory...
Plutonium represents the most anisotropic cubic element, a property not yet well
understood. Maxima occur along <111>, minima along <100>.
.... we need to invoke angular (three-body) forces, which impose difficult computational
problems. Fuchs' (1936) extended Wigner-Seitz calculations showed that for face-centered
cubic lattices, considering only the electron-electron-ion electrostatic terms, one obtains
A = 9.0, thus providing a promising possible explanation for δ-plutonium's high elastic
anisotropy. ([los])
———————————————————————————————————————
Structural mechanics delivers analogy for
metallic bonding:
On steel space frames and their nodal
joints
Sources: Tetrasteel, Italy: http://www.tetrasteel.com/eng/steel_domes.html
Sketch: Hindustan Alcox Limited. http://www.alcox.in/space_frames.html
Here metallic bonding is thought of as due to solid bars that consist of hydrogen atoms,
i. e. due to elastic H-bridges, where hydrogen’s serve for bonding. For stability
considerations this model is superior because stability can be explained due to macroscopic
stability criteria.
Then instead of unit cells that show distribution of atoms only, the true structural element
is considered.
30
Solid structures of single or double H-bridges build up the supporting structure (and not the
interplay of repulsive and attractive distant electrostatic forces between electrons and
ionic residual atoms).
For example β-bcc structures possess square bipyramids and irregular tetrahedra as
structural elements, whereas fcc structures possess octahedrons and regular tetrahedra
(both Platonic solids) as the more stable structural elements. See figures.
Determining factors: nodal joints, boundary conditions
Above beam Eigen frequencies relate to pinned-pinned boundary conditions regarding the
nodal joints of the lattice. Boundary conditions could be fixed-fixed or supports are elastic
and have displacements. Then things get complicated. But as a rule of thumb Eigen
frequencies depend on ω ∼ (nr/L)2
Because of their great masses nodal joints undergo only small vibrations. For mono-atomic
lattices, 197Au for example, we can expect harmonic vibrations, for many isotope metals
(silver, a mix of 107Ag, 109Ag), vibrations are unharmonic.
Atomic mass A as determining factor for bond lengths
The greater A, the smaller the effective bond length. Effective
bar lengths are not exactly known.
For the relation Be vs. Co the mass ratio is 6.55, therefore the
length reduction due to node volume is significant and shows
that lattice spacing a0 is not the exact parameter for specific heat.
See the paper on specific heat.
L
node
effective rod length
———————————————————————————————————
Phase transitions of alkali metals: evidence for QM’s dead end
At a pressure of 65 GPa Na is found to undergo a structural phase transition
from a bcc to a fcc modification.
Obviously, the bcc structure is mechanically unstable at high pressures.
Sisters V. F. and O. Degtyareva ascertained in their paper Structure stability in the
simple element sodium under pressure the need of a way out of QM’s dead end.
A comparison of the total energies of bcc, fcc and hpc virtual lattice structures shows very similar
results. This is not surprising because the dominant factors are electrostatic effects.
The Madelung constant determines these effects and the Madelung constants of bcc, fcc and hcp
are nearly identical! (α ≈1,7918)...
Thus, the stabilization of the bcc in alkali metals at ambient conditions can be understood as
defined by the dominant electrostatic (or Madelung) contribution to the total energy of the crystal
structure.
First-principle calculations show that all these close-packed structures have very similar energies
with the difference less than 1–5 meV ion–1, which is about equal to the accuracy of the
calculations. This makes it impossible to say why one of these structures should be more favorable
than the other.
Therefore QM is in search of explanatory factors for stability.
See [deg]
31
Phase transitions of alkali metals continued
See the problems arising in the abstract of [xie]:
Abstract
The electronic, dynamical and elastic properties ...bcc alkali metals... at high pressure are
extensively studied to reveal the origin of the phase transition from bcc...to fcc by using ab
initio calculations. The calculated 3D Fermi surface (FS) shows an anisotropic
deformation by touching the Brillion zone boundary at the N point with pressure for Li, K,
Rb and Cs due to the s→p and s→d charge transfers, respectively.
However, no clear charge transfer is found in Na, in favor of an isotropic FS even at very
high pressure. The traditional charge transfer picture and the newly proposed Home–
Rothery model in understanding the bcc→fcc transition are thus questionable in view of
their difficulties in Na. In this paper, a universal feature of pressure-induced instability of
the tetragonal shear elastic constant C ' and the softening of the transverse acoustic
phonons along the [0ξξ]-direction near the zone center for all the alkali metals is
identified. Analysis of the total energy results suggests that C ' instability associated with
the soft mode is the driving force for the bcc→fcc transition, which could be well
characterized by the tetragonal Bain's path.
Regarding the bonds of metal lattices as genuine solid bonds (and not due to electrostatic
distant forces), pressure can transform bcc structure into fcc structure.
The alternative to the QM virtual lattice due to distant electrostatic forces is the proposed genuine
lattice.
Under pressure a α-bbc structure like Na undergoes a transformation to fcc structure. This
is a macroscopic proof that fcc structures are more rigid than bcc structures ones and that
total energies of crystal structures are meaningless!
Under ambient conditions Li has bcc structure. At pressures near 40 Gpa Li transforms
from a face-centred cubic phase, via an intermediate rhombohedral phase, to a cubic
polymorph with 16 atoms per unit cell called cI16 (see figure below).
This structure can be viewed as a 2 x 2 x 2 super cell of a body-centred cubic structure but
with the atoms displaced diagonally
Sodium Na: α-bcc to → fcc crystal structure transition under pressure
Transformations of Na crystal structures
Under ambient pressure and temperature Na is α-bcc structured. Increasing pressure
changes the structure into fcc, which begins to be manifest at 65 GPa. Over 108 GPa the
fcc structure gets completely transformed into c/16. The characteristic of c/16 structure is
distortion. Transformation of fcc into distorted c/16 transforms the conductor Na into a
dielectric.
32
Table of Na crystal structure at 300K under pressure, data according to [deg]
Na
Fcc
crystal
c/16
α
oP8 ortho-
hP4
rhombic
trans-
structure
parent
bcc
space-filling-parallel-epiped
Pressure GPa
ambient
> 65
> 108
> 119
lattice [Å]
a0 = 4.2908
a* = 3.72
Ac/16 =
aorr = 4.76
∼ 4.29×√3/2
5.46
Building block
bipyramids
Octa-+2 tetrah.
Bond length [Å]
(a0/2)√3=3.71
(a*/2)√2 =2.63
V/V0
1
0.311
symmetry order
16
bonding
==
||
Bulk mod.
6.3
380? (Rh)
Shear mod. GPa
3.3
E- modulus GPa
8.83 - 10
conduction
conductor
—>
0.258
> 200
0.239
150 ? (Rh),
eeexpexpec
380?
like Rh;
expected
conductor
dielectric
insulator
Sm shows also rhomb. structure: a0= 7.8, α 22.93 ˚
Reduction of lattice constant a0 → a*. a* ∼ a0 √3/2, volume reduction ∼ 1/3.
Reduction of lattice constant of Na = 4.29 Å to 3.72 Å.
Rh possesses a lattice constant of 3.8 Å!
fcc-Na with its double H bonds (||) possesses physical properties like
rhodium, Rh, that possesses also double H bonds. So it is possible to say that Rh exists
as Rh-103 and Rh-23.
Rh-23 is the compressed Na-23.
α-bcc → fcc transition: deconstruction of the weak alkali lattice with its
large edges composed of 2 H’s in series (= =) and reconstruction of
a fcc structure with strong short edges composed of 2 H’s parallel (||).
Curent theory treats the bcc → fcc transition as a transition of the unit cell only, called
Bain deformation. No solid lattices are considered. If there is a transformation of a bcc
lattice into an fcc lattice then this transformation must be intelligible!
What happens during pressure with the α-bcc lattice bars?
A reconstruction takes place. Na (A = 23) at ambient pressure possesses α-bcc structure
with two hydrogen’s in series (= =) as bonding structures.
The content of a Na unit cell is two atoms, therefore 2×23 = 46 H’s.
The content of the new fcc unit cell is 4 atoms, therefore 4×23 = 92 H’s
Therefore 2 Na-unit cells must be compressed to form one unit cell of the new fcc
structure.
33
The Na unit cell contains 8 rods, each composed of 4 H’s. Result 32 H’s for Na rods.
(2A - 4×8)/2 = 7 hydrogen atoms are in the Na lattice nodes assembled.
Then for pressure > 65 GPa:
For the formed fcc structure with 2 H’s parallel as bonds (||) the number of hydrogen’s for
bonds is 2×24 = 48. A node contains ((4×23) - 48)/4 = 11 H’s in the fcc nodes. (For
formulas used see above.)
Balance:
Na-23
2 Na
Fcc
Fcc

unit cell
unit
unit cell
unit cell
cells
64
48
Bonding = =
Bonding =
8 rods; 4×8 = 32
24 rods; 2×24 = 48
Nodes 2×7 = 14
28
Nodes 4×11 = 44
44
Sum: hydrogen’s
92
92
Result: Some H’s that serves for Na bonding are after pressure parts of the fcc nodes.
Groups and connections of crystal structures
(See the article Element Formation)
The basic physical property is crystal structure. Crystal structures govern electrical and
heat conduction, rigidity, etc. Of course, physical properties don’t coincide strictly with
chemical properties but for some relations they do. The first table shows periodicities and
genesis of bcc metals. Note that there are 4 possible bond structures: The bars of the lattice
can be:
a: One hydrogen atom (|);
b: two H’s in series ——
c: two H’s parallel
d: two H’s parallel in series = =
For some elements a possible genesis is given. For example a periodicity of the alkaline
metals can be: 37Li +816O → 1123Na; 1123Na +816O → 1939K. Here Oxygen is a building
block for the element formation.
Both, Na and K have the same α-bcc structure and have therefore the same chemical
properties. The difference between Na and K lies in the lattice nodes: Na has nodes that
consist of 15 H’s, nodes of K consist of 31 hydrogen’s.
For brevity, sub- and superscripts are not mentioned for other formations. Their sum must
agree with sub- and superscript of the new element!
 bcc metals: | ,  and —— bonds
Genesis
Li + O → Na
Na + O → K
Br + He → Rb
Rb + 3O → Cs
Xe + He → Ba
Rb + 2He → Nb
Cr + He → Fe
V + He → Mn
Alkaline m.
α-bbc
 Li |
 Na = =
 K ——
 Rb ——
 Cs ——
Alkaline earth m.
β-bbc
 Ba ——
 Ra ——
Transition metals
 V|
 Nb |
 Ta |
 Mn =
bond 
 Fe 
 W
 Cr 
 Mo 
=
34
Next table shows hexagonal and fcc structure metals
 Hexagonal metals
 Fcc metals
-- 2H’s in series
double bonds, -- 2H’s in series
Be
-- Mg
-- Sc = 2Be + Al
-- Y = Rb + He
-- Tc = Nb + He
-- La
-- Re = Ta + He
-- Ti = Ca + He
-- Zr = Sr + He
-- Hf
-- Zn = Ti + O
-- Cd = Zr + O
Co = Sc + O
 Rh = Tc +
He
 Ir = Re +
He
--Tl =Re + C
Ru = Mo + He
Os = W + He
-- Ca = Mg + O
-- Sr = Kr + He
-- Pb = Hg + He
Al = Na + He
Cu = Co + He
Ag = Rh + He
Au = Ir + He
Ni = Fe + He
Pd = Ru + He
Pt = Os + He
Next table shows elements with diamond structures, base centered orthorhombic structures and so on…
\ P = Al + He triclinic
 trigonal (rhombohedral)
In = Ag + He tetragonal
Diamond structure
C-12 = Be-8 + He
Si = Mg + He or C + O
Ge = Zn + He
Sn = Cd + He (Sn<13,2 ˚C)
α
B-11 = Li-7+He
As = Ga + He
Sb = In + He
Te = Sn + He
Hg = Pt + He
/ monoclinic
O = C + He
F = N + He
Se = Ge + He
Bi = Tl + He
 Base centred orthorhombic
Cl-35 = P + He
Br-79 = As + He
I -127= Sb + He
Ga = Cu + He
Po = Pb + He simple cubic
Noble gases
 Ne = O + He
Ar = S + He
 Kr = Se + He
 Xe= Te + He
 Rn = Po + He
S = O + O face cent.
orthorhombic
Diamond
Imagine an fcc unit cell and than add 4 atoms inside as the
figures show. Therefore the unit cell contains 8 atoms. Any of
the 4 atoms inside of the cell has 4 connections to the 4 nearest
neighbour atoms at the surface of the cell. In the figure there
are not all rods of the diamond lattice drawn. The bond length
is 1.545 Å. Graphs:followthelemur.wordpress.com/.../
Appendix
———————————————————————————————————
Raman refutes the quasi-continuous crystal vibrational spectrum of
quantum mechanics
In the article
THE THEORY OF THE VIBRATIOINS AND THE RAMAN SPECTRUM OF THE
DIAMOND LATTICE
Helen Smith in 1947 [smi] introduced Raman’s crystal lattice theory as follows:
35
Raman and his collaborators have recently (1941) put forward a new theory of lattice dynamics acccording
to which the vibrational spectrum of a crystal consists of a few discrete lines. This is in direct contradiction
to the quasi-continuous vibrational spectrum predicted by classical or quantum mechanics. On this new
theory there are eigth fundamental frequencies of vibration for diamond; the values of these frequencies are
deduced from the observed specific heat, ultra-violet absorption and Raman spectrum...
The present author agrees that crystal lattices possess Eigen frequencies that must depend
on crystal structure. But Born’s as well as Raman’s „lattices“ are not lattices because there
are no rods! Any lattice possesses rods that are connected at node points.
A Raman spectrum for diamond excited at room temperature at a wavelength of 228.9 nm
is shown in [pra]. The fundamental line is at 1333 cm-1. This vibration is due to the
interpenetrating lattices of diamond (Raman).
Krishnan [kri] clarifies:
... Raman leads in the case of diamond to the result that the atomic vibration spectrum of this crystal should
exhibit eight distinct monochromatic frequencies. Of these, the highest frequency (1,332 cm.-1 in
spectroscopic units) corresponds to the triply degenerate vibration of the two Bravais lattices of the carbon
atoms with respect to each other, this being active in the Raman effect. The other seven frequencies represent
oscillations of the layers of carbon atoms parallel to the faces of the octahedron or the cube occurring
normal or tangential to these planes with the phase reversed at each successive equivalent layer. All the
seven modes of vibration of this description are inactive in the Raman effect as fundamentals. The octaves of
these frequencies may however, appear as frequency shifts in the Raman spectrum, though with intensities
extremely small compared with that of the Raman line of frequency shift
1,332 cm-1 . Besides the octaves, various combinations of these frequencies may also appear in the Raman
spectrum.
A diamond can be treated as a musical instrument with a fundamental tone and overtones.
Lattice constant for diamond is 3.57 Å, nearest neighbor distance = 1.55 Å.
Sound velocity of diamond: longitudinal:18350; transversal 9200 m/s, 18350/9200 = 2!
References
[ram] Raman, C V, The vibration spectrum of a crystal lattice
Proc. Indian dcad. Sci. A18 237-250 (1943)
Raman: Born’s ansatz is a failure! Thermal conduction, specific heat depends on crystal
structure!
[smi] Smith, H., THE THEORY OF THE VIBRATIOINS AND THE RAMAN SPECTRUM OF THE
DIAMOND LATTICE
Philosophical Transactions of the Royal Society of London. Series A, Mathematical and
Physical Sciences Vol. 241, No. 829, Jul. 6, 1948
[pra] St. Prawer, R. J. Nemanich, Raman spectroscopy of diamond...,
Phil. Trans. R. Soc. Lond. A (2004)
[kri] Krishnan, R. S., Raman Spectrum of Diamond
Nature 155, 171-171 (10 February 1945)
[br] Helen Brand: The melting curve of copper. C10 Independent MSci.
report.2005www.ucl.ac.uk/~ucfbhbr/melting%20copper.pdf
[che] chemicalelements.com
36
[roy] http://www.roymech.co.uk/Useful_Tables/Matter/prop_metals.htm
[rau] R. Rausch. http://www.periodensystem-online.de
[per] periodictable.com, goodfellow.com, knowledgedoor.com; webelements.com
[che] http://www.chemicalelements.com
[coh] M. L. Cohen: Concepts of modelling electrons in solids: a perspective.
Handbook of materials modelling. S. Yip, ed. Springer, Dordrecht, 2005
[hil] B. Hillmeier, textbook Baustoffkunde,
http://www.bau.tu-berlin.de/uploads/media/Baustoffkunde_I_Diplom_.pdf
[kra] Robert E. Kraig, David Roundy, Marvin L. Cohen
A study of the mechanical and structural properties of polonium
Solid State Communications 129 (2004) 411–413
www.elsevier.com/locate/ssc
[meh] M.J. Mehl, B. M. Klein, D.A. Papaconstantopoulos: First principles calculations of
elastic properties of metals. Appeared in: Intermetalllic Compounds… Volume I,
Principles, J.H. Westbrook + R.L. Fleischer, eds, John Wiley and sons London 1995,
Ch 0, pp. 195-210
[min] JR Minkel: Focus: Osmium is Stiffer than Diamond,
http://physics.aps.org/story/v9/st16
[han] M. Hanfland, I. Loa, and K. Syassen
Sodium under pressure: bcc to fcc structural transition and pressure-volume relation
to 100 GPa Phys. Rev. B 65, 184109 (2002)
Abstract :...
At a pressure of 65(1) GPa Na is found to undergo a structural phase transition from a body-centered to a facecentered-cubic modification...
[th] K. P. Thakur: Mechanical behaviour of FCC and BCC metals and their stability
1985 J. Phys. F: Met. Phys. 15 2421
[xie] Y Xie, Y M Ma, T Cui, Y Li, J Qiu and G T Zou
2008 New J. Phys. 10 063022 doi:10.1088/1367-2630/10/6/063022
Origin of bcc to fcc phase transition under pressure in alkali metals
[Li] Li Li, Shao Jian-Li, Li Yan-Fang, Duan Su-Qing and Liang Jiu-Qing
Atomistic simulation of fcc bcc phase transition in single crystal Al under uniform compression
∗ Chin. Phys. B Vol. 21, No. 2 (2012) 026402a)
Institute of Theoretical Physics and Department of Physics, Shanxi University, Taiyuan
[hol] Holden/Singer: Crystals… 1960
[Far] Faraday, Philosophical Transactions, Chapt. IX-XIII on Electricity, 1835, 1838
37
[ber] Berkson, W., Fields of Force, London 1974.
[HBE] Hoddeson, Baym, Eckert: The development of the quantum-mechanical electron
theory of metals: 1928-1933. Reviews of Modern Physics, Vol. 59, No 1. January 1987
[she] Vijay B. Shenoy, Indian Institute of Science,
http://www.physics.iisc.ernet.in/~shenoy/LectureNotes/ttlec1.pdf
Shenoy rewiews the complexity of conduction electron theory…
[ski] http://en.wikipedia.org/wiki/Skin_effect
[nd] University of Notre Dame, Physics course, author? www.nd.edu/~lwerners/F5.pdf
[foell] Föll, Helmut, textbook Materialwissenschaft:
http://www.tf.uni-kiel.de/matwis/amat/mw2_ge/index.html
[macD] MacDonald D.K.C., Am aboluten Nullpunkt, München 1961.
(Translation of Near Zero.)
[brö] B. Bröcker, Atomphysik, München 1976, 6.ed. 1997, p. 46
[dor] http://e3.physik.uni-dortmund.de/~suter/Vorlesung/Festkoerperphysik_WS12/6_Baender.pdf
[sku] skullsinthestars: http://skullsinthestars.com/2009/04/13/levitation-and-diamagnetismor-leave-earnshaw-alone/
[ear] Samuel Earnshaw, “On the Nature of the Molecular Forces which Regulate the
Constitution of the Luminiferous Ether,” Trans. Camb. Phil. Soc. 7 (1842), 97-112.
[sim] Simhony, M.
http://www.epola.co.uk/
Free downloads of Simhony’s works
http://www.epola.co.uk/Simhony/PaperBack_dnld.htm
Remark: Simhony follows QM in that he assumes extra nuclear (shell) electrons.
My atomic model: proton and electron are magnetically coupled to hydrogen.
Four hydrogen’s unite to helium. Three helium atoms unite to carbon etc.
[rol] A.D. Rollett, lectureAnisotopic Elasticity
http://neon.mems.cmu.edu/rollett/27301/L4B_AnisoElasticity_19Sep07.pdf
[los] Delta-plutonium's elastic anisotropy: Another enigma providing interatomic-bonding
insights?Los Alamos Actinide Research Quaerly, 1st quarter
2005file:///Users/johannmarinsek/Desktop/Papa%20Novemb13%20Werke%20110/plutonium%20Anisotropy
.webarchive
————————————————————————————————————————
38
Appendix
I
Born, the inventor of metallic stability due to attractive and repulsive forces between
electrons and ionic atom cores
For the sake of brevity let us cite a part of the abstract that treats crystal stability
Max Born (1940). On the stability of crystal lattices. I. Mathematical
Proceedings of the Cambridge Philosophical Society, 36 , pp 160-172
On the stability of crystal lattices. I
The stability of lattices is discussed from the standpoint of the method of small vibrations...
The stability conditions are nothing but the positive definiteness of the macroscopic deformation energy, and
can be expressed in the form of inequalities for the elastic constants. A new method is explained for
calculating these as lattice sums, and this method is applied to the three monatomic lattice types assuming
central forces. In this way one obtains a simple explanation of the fact that the face-centred lattice is stable,
whereas the simple lattice is always unstable and the body-centred also except for small exponents of the
attractive forces....
II Pauling picks number of valence electrons on demand
Pauling, L. (1949). "A Resonating-Valence-Bond Theory of Metals and Intermetallic Compounds".
Proceedings of the Royal Society of London. Series A, Mathematical and Physical Sciences (1934-1990) 196 (1046):
343.
Bibcode 1949RSPSA.196..343P. doi:10.1098/rspa.1949.0032
Abstract
The resonating-valence-bond theory of metals discussed in this paper differs from the older theory in making
use of all nine stable outer orbitals of the transition metals, for occupancy by unshared electrons and for use
in bond formation; the number of valency electrons is consequently considered to be much larger for these
metals than has been hitherto accepted. The metallic orbital, an extra orbital necessary for unsynchronized
resonance of valence bonds, is considered to be the characteristic structural feature of a metal..
III Explanation of phase transitions in terms of QM:
Yanming Ma , Mikhail Eremets, Artem R. Oganov , Yu Xie, Ivan Trojan, Sergey Medvedev, Andriy O.
Lyakhov , Mario Valle & Vitali Prakapenka
Nature 458, 182-185 (12 March 2009)
Under pressure, metals exhibit increasingly shorter interatomic distances. Intuitively, this response is
expected to be accompanied by an increase in the widths of the valence and conduction bands and hence a
more pronounced free-electron-like behaviour. But at the densities that can now be achieved experimentally,
compression can be so substantial that core electrons overlap. This effect dramatically alters electronic
properties from those typically associated with simple free-electron metals such as lithium and sodium ,
leading in turn to structurally complex phases and superconductivity with a high critical temperature. But
the most intriguing prediction—that the seemingly simple metals Li and Na will transform under pressure
into insulating states, owing to pairing of alkali atoms—has yet to be experimentally confirmed. Here we
report experimental observations of a pressure-induced transformation of Na into an optically transparent
phase at ~200 GPa (corresponding to ~5.0-fold compression). Experimental and computational data identify
the new phase as a wide bandgap dielectric with a six-coordinated, highly distorted double-hexagonal closepacked structure. We attribute the emergence of this dense insulating state not to atom pairing, but to p–d
hybridizations of valence electrons and their repulsion by core electrons into the lattice interstices. We
expect that such insulating states may also form in other elements and compounds when compression is
sufficiently strong that atomic cores start to overlap strongly.
39
IV Anisotropy of E- moduli
1:
Umesh Gandhi
Toyota Research Institute, NA. Toyota Technical Center, Ann Arbor, Mi
http://www.nist.gov/mml/acmd/structural_materials/upload/Steel-elasticity-variation-study-21012_v2.pdf
Single crystal of a iron (ferrite) is anisotropic.
Ref : Mechanical behavior of materials, by Courtney
E111= 276 GPA,
E100= 129 GPA. Ratio: 2.14
2:
http://www.lanl.gov/orgs/nmt/nmtdo/AQarchive/05spring/deltapu.html
Delta-plutonium's elastic anisotropy: Another enigma providing interatomic-bonding
insights?
3: A. Hopcroft
axial stress/strain, incl. beam bending <E110> directions ("X or Y axis") E169 GPa
<E100> directions ("45° off-axis") 130 GPa ratio= 1.3
4:
http://silicon.mhopeng.ml1.net/Silicon/
the youngs modulus of Siliconhttp://silicon.mhopeng.ml1.net/Silicon/
Rollett:---http://neon.mems.cmu.edu/rollett/27301/L4B_AnisoElasticity_19Sep07.pdf
÷÷÷÷÷÷http://magnet.atp.tuwien.ac.at/download/Skriptum138_016.pdf
40